\(\dfrac{4x}{x^2+4x+3}-1=6\left(\dfrac{1}{x+3}-\dfrac{1}{2x+2}\right)\left(x\ne-3;x\ne-1\right)\\ < =>\dfrac{4x}{x^2+4x+4-1}-1=\dfrac{6}{x+3}-\dfrac{6}{2x+2}\\ < =>\dfrac{4x}{\left(x+2\right)^2-1}-1=\dfrac{6}{x+3}-\dfrac{6}{2\left(x+1\right)}\\ < =>\dfrac{4x}{\left(x+1\right)\left(x+3\right)}-1=\dfrac{6}{x+3}-\dfrac{3}{x+1}\)
suy ra:
`4x-(x+1)(x+3)=6(x+1)-3(x+3)`
\(< =>4x-\left(x^2+3x+x+3\right)=6x+6-3x-9\)
\(< =>4x-x^2-4x-3=6x+6-3x-9\)
\(< =>-x^2+4x-4x-6x+3x-3-6+9=0\)
\(< =>-x^2-3x=0\\ < =>x^2+3x=0\\ < =>x\left(x+3\right)=0\\ < =>\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\left(tm\right)\\x=-3\left(ktmđk\right)\end{matrix}\right.\)
vậy pt có tập nghiệm S={0}
